An astronaut takes a ball of mass m from Eart | vedclass

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(A) The initial total mechanical energy of the ball on the Earth's surface is $TE_i = -\frac{GM_e m}{R_e}$.Given the altitude $h = 318.5 \ km$ and $R_

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) The initial total mechanical energy of the ball on the Earth's surface is $TE_i = -\frac{GM_e m}{R_e}$.Given the altitude $h = 318.5 \ km$ and $R_e = 6370 \ km$,we find $h = \frac{R_e}{20}$.The final total mechanical energy of the ball in a circular orbit at altitude $h$ is $TE_f = -\frac{GM_e m}{2(R_e + h)}$.Substituting $h = \frac{R_e}{20}$,we get $TE_f = -\frac{GM_e m}{2(R_e + R_e/20)} = -\frac{GM_e m}{2(21R_e/20)} = -\frac{10 GM_e m}{21 R_e}$.The change in total mechanical energy is $\Delta TE = TE_f - TE_i = -\frac{10 GM_e m}{21 R_e} - (-\frac{GM_e m}{R_e})$.$\Delta TE = \frac{GM_e m}{R_e} (1 - \frac{10}{21}) = \frac{11 GM_e m}{21 R_e}$.Comparing this with $x \frac{GM_e m}{21 R_e}$,we get $x = 11$.

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